Answer:
[tex]\dfrac{y-1}{y(y+4)}[/tex]
Step-by-step explanation:
Given expression:
[tex]\implies \dfrac{2y-1}{y^2+4y}-\dfrac{y-2}{y^2+2y-8}[/tex]
Factor the denominators of both fractions:
[tex]\implies \dfrac{2y-1}{y(y+4)}-\dfrac{y-2}{y^2+4y-2y-8}[/tex]
[tex]\implies \dfrac{2y-1}{y(y+4)}-\dfrac{y-2}{y(y+4)-2(y+4)}[/tex]
[tex]\implies \dfrac{2y-1}{y(y+4)}-\dfrac{y-2}{(y-2)(y+4)}[/tex]
Cancel the common factor (y - 2) of the right fraction:
[tex]\implies \dfrac{2y-1}{y(y+4)}-\dfrac{1}{(y+4)}[/tex]
Adjust the fractions based on the LCM of y(y + 4):
[tex]\implies \dfrac{2y-1}{y(y+4)}-\dfrac{y}{y(y+4)}[/tex]
[tex]\textsf{Apply the fraction rule} \quad \dfrac{a}{c}-\dfrac{b}{c}=\dfrac{a-b}{c}:[/tex]
[tex]\implies \dfrac{2y-1-y}{y(y+4)}[/tex]
Simplify:
[tex]\implies \dfrac{y-1}{y(y+4)}[/tex]
